Question: The radius of a circle is decreasing at a rate of $6.5$ meters per minute. At a certain instant, the radius is $12$ meters. What is the rate of change of the area of the circle at that instant (in square meters per minute)? Choose 1 answer: Choose 1 answer: (Choice A) A $-288\pi$ (Choice B) B $-144\pi$ (Choice C) C $-156\pi$ (Choice D) D $-42.25\pi$
Setting up the math Let... $r(t)$ denote the radius at time $t$, and $A(t)$ denote the circle's area at time $t$. We are given that $r'(t)=-6.5$ and that $r(t_0)=12$ for a specific time $t_0$. Note that $r'(t)$ is negative because the radius is decreasing. We want to find $A'(t_0)$. Relating the measures $A(t)$ and $r(t)$ relate to each other through the formula for the area of a circle: $A(t)=\pi[r(t)]^2$ We can differentiate both sides to find an expression for $A'(t)$ : $A'(t)=2\pi r(t)r'(t)$ Using the information to solve Let's plug ${r(t_0)}={12}$ and ${r'(t_0)}={-6.5}$ into the expression for $A'(t_0)$ : $\begin{aligned} A'(t_0)&=2\pi{r(t_0)}{r'(t_0)} \\\\ &=2\pi({12})({-6.5}) \\\\ &=-156\pi \end{aligned}$ In conclusion, the rate of change of the area of the circle at that instant is $-156\pi$ square meters per minute. Since the rate of change is negative, we know that the area is decreasing.